Tangent vectors as derivations and dot product

Question

Take $\mathbb{R}^n$ with the Euclidean metric to be the manifold of interest for this question. Suppose we have two vectors

$v_p = v^i (\partial/\partial x^i)_p$ and

$w_p = w^j (\partial/\partial x^j)_p$

How is the dot product between $v_p$ and $w_p$ defined and what does it look like?

Answer 1

The Euclidean metric is the following

$$ ds^2=\sum_{k=1}^n dx_k\otimes dx_k,$$

where $n$ clearly is the dimension of $\mathbb{R}^n$.

So $ds^2(v_p,w_p)=\big(\sum_{k=1}^n dx_k\otimes dx_k\big)(v^i\partial_i,w^j\partial_j)=\sum_{k=1}^n dx_k(v^i\partial_i)dx_k(w^i\partial_j)$;

where $dx_k(v^i\partial_i)=\delta_{ik}v^i$ and $dx_k(w^j\partial_j)=\delta_{jk}w^j$.

Then you have $$ds^2(v_p,w_p)=\sum_{k=1}^n v^kw^k.$$

Answer 2

Just like on $\Bbb{R}^n$: $v_p \cdot w_p = \sum_{i = 1}^n v^i w^i$.

This comes directly from the isomorphism between $T_p M$ and $\Bbb{R}^n$ given by $v_p \mapsto (v^1, \dotsc, v^n)$.